Oh, dude, so like, capacitance is just the ratio of charge to voltage, right? So if you've got 60 coulombs of charge and 12 volts of potential difference, you just divide 60 by 12 and boom, you get 5 farads. Easy peasy lemon squeezy.
a = 20 b = 60 c = 100
If 8c = 60, then c = 60/8 = 7.5.
If we're talking strictly algebra. 12 equals p of c can be written as: 12= p(c) meaning, 12 is the answer for some function p, when c is the variable.
yes. b/c 12*5=60
t = 4/5
C = Q/V; capacitance = charge / voltage. Therefore, you would also have to know the voltage.
C = capacitance, f = frequency ===> Capacitive reactance = 1 / [ 2(pi)fC ] 663 = 1 / [ 2(pi)(60)C ] 663 x 2 x pi x 60 x C = 1 C = 1 / (663 x 2 x pi x 60) = 1 / (663 x 120 x pi) = 1 / 249,945.1 = 4 x 10-6 = 4 microfarads (almost exactly)
a = 20 b = 60 c = 100
The value of c/30 that equals 12 is 12 . In order to make that value, 'c' must be 360 .
If 8c = 60, then c = 60/8 = 7.5.
If we're talking strictly algebra. 12 equals p of c can be written as: 12= p(c) meaning, 12 is the answer for some function p, when c is the variable.
yes. b/c 12*5=60
t = 4/5
-12 + c = 19 Add 12 to both sides: c = 19+12 = 31
C
To determine the capacitance of a system, you can use the formula C Q/V, where C is the capacitance, Q is the charge stored in the system, and V is the voltage across the system. By measuring the charge and voltage, you can calculate the capacitance of the system.
space charge region in a diode or say a bjt for better understanding is same as the depletion region, both transition capacitance and depletion capacitance are the same c= (epsilon*A)/d ; where ... c is capacitance A is area and d is the depletion width the other type of capacitance is the diffusion capacitance c= (T*I)/(n*V) where ... c is the capacitance T is transition ti me I is the drift current n is emission coefficient ... its value is 1 for germanium and V is thermal voltage .. 26mv